On page 72 of PDE evans, the line after equation $(21)$ says:
$$\frac{1}{| \partial B(x,t)|}\int_{\partial B(x,t)} g(y) \, dS(y) = \frac{1}{| \partial B(0,1)|}\int_{\partial B(0,1)} g(x+tz) \, dS(z)$$
I do not understand the derivation of this one either. All I know is that they are substituting $y=x+tz$. Then you are supposted to transform $(x,t) \rightarrow (0,1)$ and $dS(y) \rightarrow dS(z)$ somehow. But these are where I'm stuck.
The linear transformation $L(z) = x+tz$ has the following properties:
$L(0)=x$
$|L(a)-L(b)|=t|a-b|$ for all $a,b$
As a consequence, $L(\partial B(0,1))=\partial B(x,t)$. Also, since the transformation scales all distances by the factor of $t$, it scales the $k$-dimensional measure of sets by $t^k$. In particular, Lastly, $dS(y)=t^{n-1} dS(z)$. This factor of $t^{n-1}$ cancels out because $$|\partial B(x,t)|=t^{n-1}|\partial B(0,1)|$$